Problem: Graph this system of equations and solve. $-9x-5y = -20$ $8x+10y = -10$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
Solution: Convert the first equation, $-9x-5y = -20$ , to slope-intercept form. $y = -\dfrac{9}{5} x + 4$ The y-intercept for the first equation is $4$ , so the first line must pass through the point $(0, 4)$ The slope for the first equation is $-\dfrac{9}{5}$ . Remember that the slope tells you rise over run. So in this case for every $9$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $9$ positions down from $(0, 4)$ is $(5, -5)$ Graph the blue line so it passes through $(0, 4)$ and $(5, -5)$ Convert the second equation, $8x+10y = -10$ , to slope-intercept form. $y = -\dfrac{4}{5} x - 1$ The y-intercept for the second equation is $-1$ , so the second line must pass through the point $(0, -1)$ The slope for the second equation is $-\dfrac{4}{5}$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $4$ positions down from $(0, -1)$ is $(5, -5)$ Graph the green line so it passes through $(0, -1)$ and $(5, -5)$ The solution is the point where the two lines intersect. The lines intersect at $(5, -5)$.